Chemistry Homework Answer Key #2
Chapter 8 - Spontaneous Processes and Thermodynamic
Equilibrium
23. For the reaction,
Substance | So (J K-1) |
N2H4 (l) | 121.21 |
O2 (g) | 205.03 |
NO2 (g) | 239.95 |
H2O (l) | 69.91 |
27. Since the value of DSuniverse (DSuniverse = DSsystem + DSsurroundings) is positive for all spontaneous processes, the change in entropy of the surroundings must be greater in magnitude and opposite in sign than the DS for the reaction. That is, DSsurroundings > + 44.7 J K-1 mol-1.![]()
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If hydrazine were in the gaseous state instead of the liquid state, its standard entropy would be higher, i.e., more positive. Thus, the total entropy of the reactants would be greater, leading to an even more negative value for the DS° of the reaction.
31.
a)
b)![]()
c) No. Since the value of DG is positive, this is not a spontaneous process.
d) At equilibrium, DG equals zero. Therefore,
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This calculation provides an estimate of the melting point of solid ammonia. The actual melting point of ammonia is 195.45 K. Not a bad estimate (they don't all work out so well).
35. To consume the three mol of oxygen liberated in
the first reaction, three mol of carbon must be used. Thus, the coefficients
of the second reaction listed, as well as its corresponding DG,
need to be multiplied by three. The DG for the
combined system of reactions is given by the sum of the individual reaction
DG
values, as shown below:
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37. The values of DH°
and DS° for the
following reactions were obtained using the thermodynamic data listed in
Appendix D.
a) 4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s) DH° = -1648.4 kJ DS° = -549.41 J K-147. The answer to problem 22 was shown on Answer Key 1 and is repeated below:With a negative value of both DH° and DS° , this reaction will proceed spontaneously only at low temperature. To calculate, the same approach that was outlined in #31 is used.
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Thus, the reaction will be spontaneous when the temperature is less than 3000 K.
b) SO2 (g) + ½ O2 (g) ® SO3 (g) DH° = -98.89 kJ DS° = -93.98 J K-1
As in part a, this reaction will proceed spontaneously only at low temperature (less than 1050 K).
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In the above scenario, the iron is cooled rapidly, from 100 to 0ºC. If we were to perform the cooling in "baby steps", the process would begin to approach a reversible change, i.e., the total entropy change of the universe would approach zero.![]()
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a) In two steps, the entropy is given by:
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The change in entropy for the iron is unchanged. But, examining the change in entropy for the water, we have.
For the first step the temperature drops from 100 to 50ºC.
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The change in entropy for the first step is:
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Likewise, for the second step:
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The change in entropy for the second step is:
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So, the total change in entropy for the water is +8.49 J K-1. The change in entropy for the universe is 0.66 J K-1, a smaller increase than was seen for the single step process.
b) Repeating the procedure with four steps decreases DSuniverse even more. An abbreviated treatment follows. DS for the iron remains the same (-7.83 J K-1), as was the case above. For the water, the heat gained occurs in four equivalent transfers, each of 627.5 J.
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So the DSuniverse = +0.32 J K-1.
51. Since the absolute entropy is equal to the integral
of Cp/T vs. 1/T from 0 to 298 K, the curve with
the greatest area under the curve will have the highest entropy. Therefore,
gold would be expected to have the highest entropy, and it does (S°Au
= 47.40 J K-1, S°Cu
= 33.15 J K-1).
53. For the reaction, S (s, rhombic) ® S (s, monoclinic) DS° = +0.8 J K-1
Given the fact that this transition is non-spontaneous at low temperature, the positive change in entropy means that DH must also be positive. Therefore DH (not DH° ) is equal to +400 J/mol. To find the value of DS at the temperature of the transition, the following calculation is performed.
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55. For the reaction,
3 CO2 (g) + Si3N4 (s) ® 3 SiO2 (s, quartz) + 2 N2 (g) + 3 C (s, graphite)
Substance | DGof (kJ) |
CO2 (g) | -394.36 |
Si3N4 (s) | -642.6 |
SiO2 (s, quartz) | -856.67 |
N2 (g) | 0 |
C (s, graphite) | 0 |
56. Using the cis form as an example, the change
in enthalpy of formation, DH°f,
and Gibbs free energy of formation, DG°f,
correspond to the following reaction:
Pt (s) + N2 (g) + 3 H2 (g) + I2 (s) ®cis-Pt(NH3)2I2 DH°f = -286.56 kJ DG°f = -130.25 kJ
The change in entropy for this reaction is given by:
As with any reaction,
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By the same process, S°(trans-Pt(NH3)2I2) = 219.37 J K-1.
57.
a) The values of DH° and DS° for the following reaction were obtained using the thermodynamic data listed in Appendix D.
2 CuCl2 (s) ® 2 CuCl (s) + Cl2 (g) DH° = +165.80 kJ DS° = +179.22 J/K
b) Using these data, an estimate for DG at 590 K can be made.DG = DH - TDS = 165.80 kJ - (590 K)(0.17922 J K-1) = + 60.1 kJ
c) Using the measured values of DH590 and DS590, a refined calculation of DG590 can be performed.
DG590 = DH590 - TDS590 = 158.36 kJ - (590 K)(0.17774 J K-1) = + 53.5.1 kJ
The original estimate was in error by +12%.